Question

Suppose $l, m, n$ respectively represent the coefficient of $x^{10}$, the constant term and the coefficient of $x^{-10}$ in the expansion of $\left(a x^{2}+\frac{b}{x^{3}}\right)^{15}$.If $\frac{l}{m}+\frac{m}{n}=\frac{26}{11}$, then $a^{2}: b^{2}=$

• A. $16: 9$
• B. $9: 4$
• C. $4: 1$
• D. $1: 25$

Answer 1

Answer: (B)

Given binomial $\left(a x^{2}+\frac{b}{x^{3}}\right)^{15}$,
so the general term, $T_{r+1}={ }^{15} C_{r} a^{15-r} b^{r} x^{30-5 r}$
$\therefore $ The coefficient of $x^{10}$ is ( at $r=4$ )
$={ }^{15} C_{u} a^{11} b^{4}=$ (given)
Similarly, $m=$ the constant term (at $r=6$ )
$={ }^{15} C_{6} a^{9} b^{6}$
and $n=$ coefficient of $x^{-10}$ is $($ at $r=8)={ }^{15} C_{8} a^{7} b^{8}$
$\because \frac{l}{m}+\frac{m}{n}=\frac{26}{11}$
$\Rightarrow \frac{{ }^{15} C_{4} a^{11} b^{4}}{{ }^{15} C_{6} a^{9} b^{6}}+\frac{{ }^{15} C_{6} a^{9} b^{6}}{{ }^{15} C_{8} a^{7} b^{8}}=\frac{26}{11}$
$\Rightarrow \frac{6 ! 9 ! a^{2}}{4 ! 111 b^{2}}+\frac{8 ! 7 ! a^{2}}{6 ! 9 ! b^{2}}=\frac{26}{11}$
$\Rightarrow \frac{6 \times 5}{11 \times 10}\left(\frac{a^{2}}{b^{2}}\right)+\frac{7}{9}\left(\frac{a^{2}}{b^{2}}\right)=\frac{26}{11} $
$\Rightarrow \frac{a^{2}}{b^{2}}\left(\frac{3}{11}+\frac{7}{9}\right)=\frac{26}{11}$
$\Rightarrow \frac{a^{2}}{b^{2}}\left(\frac{27+77}{99}\right)=\frac{26}{11}$
$ \Rightarrow \frac{a^{2}}{b^{2}}=\frac{9 \times 26}{104}=\frac{9}{4}$